A355342 - OEIS

%I #8 Jul 24 2022 03:57:32

%S 0,1,-2,-1,3,0,1,-4,2,0,-1,5,-5,0,0,1,-6,9,-2,0,0,-1,7,-14,7,0,0,0,1,

%T -8,20,-16,2,0,0,0,-1,9,-27,30,-9,0,0,0,0,1,-10,35,-50,25,-2,0,0,0,0,

%U -1,11,-44,77,-55,11,0,0,0,0,0,1,-12,54,-112,105,-36,2,0,0,0,0,0,-1,13,-65,156,-182,91,-13,0,0,0,0,0,0,1,-14,77,-210,294,-196,49,-2,0,0,0,0,0,0,-1,15,-90,275,-450,378,-140,15,0,0,0,0,0,0,0

%N G.f.: A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * C(x)^n, where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

%H Paul D. Hanna, <a href="/A355342/b355342.txt">Table of n, a(n) for n = 0..1275</a>

%F G.f. A(x) = Sum_{n>=0} a(n) * x^n is equal to the following expressions; here, C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

%F (1) A(x) = -1/C(x) * Product_{n>=1} (1 - x^n/C(x)) * (1 - x^(n-1)*C(x)) * (1-x^n), by the Jacobi triple product identity.

%F (2) A(x) = Sum_{n=-oo..+oo} (-1)^n * x^(n*(n+1)/2) * C(x)^n.

%F (3) A(x) = Sum_{n>=0} (-1)^n * x^(n*(n+1)/2) * (C(x)^n - 1/C(x)^(n+1)).

%F (4) A(x) = 1 - Sum_{n>=0} x^(n*(n+1)/2) * ( [y^n] (1 + 2*y*x)/(1+x + y*x^2) ).

%F (5) A(x) = 1 - Sum_{n>=1} (-1)^n * x^(n*(n-1)/2) * Sum_{k=0..n} A244422(n,k) * x^k.

%e G.f.: A(x) = x - 2*x^2 - x^3 + 3*x^4 + x^6 - 4*x^7 + 2*x^8 - x^10 + 5*x^11 - 5*x^12 + x^15 - 6*x^16 + 9*x^17 - 2*x^18 - x^21 + 7*x^22 - 14*x^23 + 7*x^24 + x^28 - 8*x^29 + 20*x^30 - 16*x^31 + 2*x^32 - x^36 + 9*x^37 - 27*x^38 + 30*x^39 - 9*x^40 + x^45 - 10*x^46 + 35*x^47 - 50*x^48 + 25*x^49 - 2*x^50 + ...

%e such that

%e A(x) = ... + x^6/C(x)^4 - x^3/C(x)^3 + x/C(x)^2 - 1/C(x) + 1 - x*C(x) + x^3*C(x)^2 - x^6*C(x)^3 + x^10*C(x)^4 +- ...

%e where

%e C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + 4862*x^9 + 16796*x^10 + ... + A000108(n)*x^n + ...

%e The coefficients of x^k in (-1)^n * x^(n*(n+1)/2) * (C(x)^n - 1/C(x)^(n+1)) begin:

%e n = 0: [0, 1,  1,  2,  5,  14,  42,  132,  429,  1430,  4862,  16796, ...];

%e n = 1: [0, 0, -3, -3, -7, -19, -56, -174, -561, -1859, -6292, -21658, ...];

%e n = 2: [0, 0,  0,  0,  5,   5,  15,   45,  141,   457,  1520,   5159, ...];

%e n = 3: [0, 0,  0,  0,  0,   0,   0,   -7,   -7,   -28,   -91,   -301,  ...];

%e n = 4: [0, 0,  0,  0,  0,   0,   0,    0,    0,     0,     0,      9, ...]; ...

%e forming a table the column sums of which yield this sequence.

%e The g.f. may also be written as

%e A(x) = 0 + (-2*x + 1)*x - (-3*x + 1)*x^3 + (2*x^2 - 4*x + 1)*x^6 - (5*x^2 - 5*x + 1)*x^10 + (-2*x^3 + 9*x^2 - 6*x + 1)*x^15 - (-7*x^3 + 14*x^2 - 7*x + 1)*x^21 + (2*x^4 - 16*x^3 + 20*x^2 - 8*x + 1)*x^28 - (9*x^4 - 30*x^3 + 27*x^2 - 9*x + 1)*x^36 + (-2*x^5 + 25*x^4 - 50*x^3 + 35*x^2 - 10*x + 1)*x^45 + ...

%e compare to

%e (1 + 2*y*x)/(1+x + y*x^2) = 1 - (-2*y + 1)*x + (-3*y + 1)*x^2 - (2*y^2 - 4*y + 1)*x^3 + (5*y^2 - 5*y + 1)*x^4 - (-2*y^3 + 9*y^2 - 6*y + 1)*x^5 + (-7*y^3 + 14*y^2 - 7*y + 1)*x^6 - (2*y^4 - 16*y^3 + 20*y^2 - 8*y + 1)*x^7 + (9*y^4 - 30*y^3 + 27*y^2 - 9*y + 1)*x^8 - (-2*y^5 + 25*y^4 - 50*y^3 + 35*y^2 - 10*y + 1)*x^9 + ...

%e The terms of this sequence may be written as a triangle:

%e 0,

%e 1, -2,

%e -1, 3, 0,

%e 1, -4, 2, 0,

%e -1, 5, -5, 0, 0,

%e 1, -6, 9, -2, 0, 0,

%e -1, 7, -14, 7, 0, 0, 0,

%e 1, -8, 20, -16, 2, 0, 0, 0,

%e -1, 9, -27, 30, -9, 0, 0, 0, 0,

%e 1, -10, 35, -50, 25, -2, 0, 0, 0, 0,

%e -1, 11, -44, 77, -55, 11, 0, 0, 0, 0, 0,

%e 1, -12, 54, -112, 105, -36, 2, 0, 0, 0, 0, 0,

%e -1, 13, -65, 156, -182, 91, -13, 0, 0, 0, 0, 0, 0,

%e 1, -14, 77, -210, 294, -196, 49, -2, 0, 0, 0, 0, 0, 0,

%e -1, 15, -90, 275, -450, 378, -140, 15, 0, 0, 0, 0, 0, 0, 0,

%e 1, -16, 104, -352, 660, -672, 336, -64, 2, 0, 0, 0, 0, 0, 0, 0,

%e ...

%o (PARI) {a(n) = my(A,C = serreverse(x-x^2 +x^2*O(x^n))/x);

%o A = sum(m=-n-1,n+1, (-1)^m * x^(m*(m+1)/2) * C^m); polcoeff(A,n)}

%o for(n=0,70,print1(a(n),", "))

%o (PARI) {a(n) = my(A,C = serreverse(x-x^2 +x^2*O(x^n))/x, M = sqrtint(2*n+9));

%o A = sum(m=0,M, (-1)^m * x^(m*(m+1)/2) * (C^m - 1/C^(m+1))); polcoeff(A,n)}

%o for(n=0,70,print1(a(n),", "))

%Y Cf. A244422, A355341, A355343.

%Y Cf. A034807.

%K sign

%O 0,3

%A _Paul D. Hanna_, Jul 22 2022